HARD · MOTION IN A STRAIGHT LINE
STOPPING DISTANCE WITH REACTION TIME
A driver is travelling at 30 m/s when they spot a hazard. Their reaction time is 0.8 s, after which the brakes apply a constant deceleration of 6 m/s². Find the total stopping distance from the moment the hazard is seen.
§ 01
Step-by-step solution
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Step 1
During the reaction time t_r, the car moves at constant velocity v₀. Find the reaction distance d_react.
Hint
No braking happens during reaction time — the car is still travelling at full speed.
Step 2
Find the time t_brake it takes for the car to decelerate from v₀ to rest under constant deceleration a_brake.
Step 3
Find the braking distance d_brake — the distance covered from the moment brakes are applied until the car stops.
Step 4
Sum the reaction distance and the braking distance to get the total stopping distance d_total.
Solution walkthrough
Stopping a car in a real emergency involves two distinct phases that must be handled separately. During the reaction time (0.8 s), the driver has not yet acted — the car travels at full speed. Reaction distance: d_react = 30 × 0.8 = 24 m. Once the brakes engage, the car decelerates at 6 m/s². The braking time is found by asking when velocity reaches zero: 0 = 30 − 6·t, so t_brake = 5 s. The braking distance is d_brake = v₀²/(2·a) = 900/12 = 75 m. (This formula comes from v_f² = v₀² + 2·a·d; setting v_f = 0 and solving for d.) Total stopping distance: 24 + 75 = 99 m. The important insight is that reaction time is not trivial — at 30 m/s, the driver travels 24 m before the brakes even begin. That is nearly a quarter of the total stopping distance, and it scales linearly with speed, while braking distance scales with v².
§ 02
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